Skip to main content

Full text of "Analytical Mechanics"

See other formats

1. Supposing the weights in Fig. 98 to be in equilibrium and the contacts to be smooth, find the relation between the two weights.
If Wi is given a virtual displacement towards the left along the inclined plane, then the virtual work is ~Tds + Wi  ds sin a + JV  0 = 0,
T =Wi sin a. T=W*
Wz = Wi sin a.
FIG. 98.
2. Two uniform rods of equal weight W and equal length a are jointed at one end and placed, as shown in Fig. 99, in a vertical plane on a smooth horizontal table. A string of length Z joins the middle points of the rods. Find the tension of the string.
The following forces act upon each rod  the weight of the rod, the pull of the string, the reaction at the joint, and the reaction of the table. Suppose a slight displacement to be given to the system by pressing downward at the joint. The work done by the force which produced the displacement equals the sum of the work done by the other forces which act upon the rods during the displacement. But since both the force applied and the displacement produced are very small their product is negligible. Therefore the sum of the work done by all the other forces is zero.
The reactions at the ends of the rods do not contribute to the virtual work because each of the reactions is perpendicular to the corresponding
surface of contact along which the displacement takes place. Therefore the weights and the tensile force of the string contribute all the virtual work. If dl and dh denote, respectively, the increase in length of the string and the distance through which the centers of mass of the rods are lowered during the virtual displacement the virtual work takes the form
But from the figure I = a sin 6, and h = a cos 6.   Therefore dl = a cos 6 dd and dh =  a sin 6 dd.   Making these substitutions and simplifying we